Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in JAVA):

Solution:
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF587C extends PrintWriter {
	CF587C() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF587C o = new CF587C(); o.main(); o.flush();
	}

	static class V {
		int i, c;
		V(int i, int c) { this.i = i; this.c = c; }
	}
	int[] next, jj; int l_ = 1;
	int link(int l, int j) { next[l_] = l; jj[l_] = j; return l_++; }
	int[] ao, dd;
	int[][] pp;
	int[][][] id;
	int ln;
	void init(int n, int m) {
		next = new int[1 + m * 2];
		jj = new int[1 + m * 2];
		ao = new int[n];
		dd = new int[n];
		ln = 0;
		while (1 << ln + 1 <= n - 1)
			ln++;
		pp = new int[n][ln + 1];
		id = new int[n][ln + 1][10];
	}
	void copy(int[] aa, int[] bb) {
		for (int i = 0; i < 10; i++)
			bb[i] = aa[i];
	}
	void merge(int[] aa, int[] bb, int[] cc) {
		int i = 0, j = 0, k = 0;
		while (k < 10)
			if (aa[i] < bb[j])
				cc[k++] = aa[i++];
			else
				cc[k++] = bb[j++];
	}
	void dfs(int p, int i, int d) {
		dd[i] = d;
		pp[i][0] = p;
		for (int k = 1; 1 << k <= d; k++) {
			int q = pp[i][k - 1];
			pp[i][k] = pp[q][k - 1];
			merge(id[i][k - 1], id[q][k - 1], id[i][k]);
		}
		for (int l = ao[i]; l != 0; l = next[l]) {
			int j = jj[l];
			if (j != p)
				dfs(i, j, d + 1);
		}
	}
	int[] ans = new int[10];
	int[] tmp = new int[10];
	void query(int i, int j) {
		if (dd[i] < dd[j]) {
			int tmp = i; i = j; j = tmp;
		}
		for (int k = ln; k >= 0; k--)
			if (1 << k <= dd[i] - dd[j]) {
				copy(ans, tmp);
				merge(tmp, id[i][k], ans);
				i = pp[i][k];
			}
		if (i == j) {
			copy(ans, tmp);
			merge(tmp, id[j][0], ans);
			return;
		}
		for (int k = ln; k >= 0; k--)
			if (1 << k <= dd[i] && pp[i][k] != pp[j][k]) {
				merge(ans, id[i][k], tmp);
				merge(tmp, id[j][k], ans);
				i = pp[i][k];
				j = pp[j][k];
			}
		merge(ans, id[i][0], tmp);
		merge(tmp, id[j][0], ans);
		copy(ans, tmp);
		merge(tmp, id[pp[i][0]][0], ans);
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int q = sc.nextInt();
		init(n, n - 1);
		for (int h = 0; h < n - 1; h++) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			ao[i] = link(ao[i], j);
			ao[j] = link(ao[j], i);
		}
		V[] vv = new V[m];
		for (int h = 0; h < m; h++) {
			int c = sc.nextInt() - 1;
			vv[h] = new V(h, c);
		}
		Arrays.sort(vv, (u, v) -> u.c != v.c ? u.c - v.c : u.i - v.i);
		for (int i = 0; i < n; i++)
			Arrays.fill(id[i][0], m);
		for (int i = 0, j; i < m; i = j) {
			int c = vv[i].c;
			j = i + 1;
			while (j < m && vv[j].c == c)
				j++;
			for (int h = 0; h < 10 && i + h < j; h++)
				id[c][0][h] = vv[i + h].i;
		}
		dfs(-1, 0, 0);
		while (q-- > 0) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			int k = sc.nextInt();
			Arrays.fill(ans, m);
			query(i, j);
			while (k > 0 && ans[k - 1] == m)
				k--;
			print(k);
			for (int h = 0; h < k; h++)
				print(" " + (ans[h] + 1));
			println();
		}
	}
}